PROBABILITY Problem 1. (25 points) An Integrated Circuit (IC) chip is declared a
ID: 3365397 • Letter: P
Question
PROBABILITY
Explanation / Answer
Let probability of 1st chip being good =P(A) = 3/5, P(A') = 1 - 3/5 = 2/5
Let probability of 2nd chip being good =P(B)
Given P(B/A) = 4/5 and P(B/A') = 2/5
P(B) = P(B/A) * P(A) + P(B/A') * P(A) = (4/5 * 3/5) + (2/5 * 3/5) = 12/25 + 6/25 = 18/25
Therefore P(B') = 1 - 18/25 = 7/25
a) Probability that both chips are good = P(A) * P(B) = 3/5 * 18/25 = 54/125
b) Probability that both chips are bad = P(A') * P(B') = 2/5 * 7/25 = 14/125
c) Probability that exactly one chip is good = (1st is good, 2nd is bad) + P(1st is bad, 2nd is good)
= P(A) * P(B') + P(A') * P(B) = (3/5 * 7/25) + (2/5 * 18/25) = 21/125 + 36/125 = 57/125
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